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Let $\sigma_0,\sigma_x,\sigma_y,\sigma_z$ stand for the $2\times 2$ identity matrix and the well known pauli matrices:

\begin{equation} \sigma_0=\left(\begin{array}{cc}1&0\\0&1\end{array}\right), \sigma_x=\left(\begin{array}{cc}0&1\\1&0\end{array}\right), \sigma_y=\left(\begin{array}{cc}0&-i\\i&0\end{array}\right), \sigma_z=\left(\begin{array}{cc}1&0\\0&-1\end{array}\right) \end{equation}

How would you prove that the $4\times 4$ Hermitian matrices constitute a linear vector space with basis the tensor products of $\sigma_0,\sigma_x,\sigma_y,\sigma_z$?

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It is not a homework problem. It is an assertion on a textbook which is left unproven. –  Euclean May 28 '12 at 11:57
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1 Answer 1

up vote 2 down vote accepted

I am not sure whether I understood the question correctly. First of all any $2\times2$ Hermitian matrix $A$ can be written as a real linear combination of the Pauli matrices and they form the basis. To see this, notice that the inner product is given by the trace, and the coefficient $x_s$ of $\sigma_s$ will be $\mathrm{Tr}(A\cdot\sigma_s)$ where $s=0,x,y,z$.

After this you can simply use the fact that for two vector spaces $V_1,~V_2$ with basis $\{e_i\}$ and $\{f_j\}$ respectively, the tensor product $V_1\otimes V_2$ is a vector space with basis elements are given by $\{e_i\otimes f_j\}$.

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